Algebraic realizations of cyclic isometries of the rational K3 lattice

We prove: For all n and all n-cylic isometriesφ : ΛQ → ΛQ of the rational K3 lattice there exists an algebraic realization of φ, i.e. marked algebraic K3 surfaces (S,ηS) and (M,ηM), whereM is a moduli space of sheaves on S, and a Hodge isometry ψ : H2(S,Q)→ H2(M,Q) such that φ = ηM ◦ψ ◦ η−1 S . 1 Facts about cyclic isometries We need the existence of a triple ((S,ηS), (M,ηM),ψ) to have a point in our moduli space from which to start deforming. The moduli space will be seen to be covered by “twistor lines” and deformations of our examples along these lines will be described in a later talk. We begin by recalling our essential definition. Definition 1.1. Let L1 and L2 be lattices. An isometry φ : L1 ⊗Z Q → L2 ⊗Z Q is called n-cyclic if L1 φ−1(L2) ∩ L1 ∼ = Z/nZ By O(L) we denote the (integral) isometry group of L and by O(LQ) the (rational) isometry group of LQ. By Λwe denote the K3 lattice U⊕3 ⊕ (−E8). Definition 1.2. For any φ ∈ O(ΛQ) of n-cyclic type its double orbit [φ] is given by O(Λ)φO(Λ) ⊆ O(ΛQ). The following result says that all n-cyclic isometries of the K3 lattice are conjugate to each other using integral isometries. This will be helpful later, since we can just produce a Hodge isometry associated to some n-cyclic isometry of the K3 lattice (not necessarily the the one we started with), and then conjugate it from there. Proposition 1.3 (Proposition 3.3 in [Bus15]). Let φ1 and φ2 be rational isometries of ΛQ of n-cyclic type. Then [φ1] = [φ2]. The proof of Proposition 1.3 is lattice-theoretic and very technical. We will skip it.